The mass of the ball is 0.050kg, and at the instant shown the value of the tension is 4N. Find the net force on the ball and find the acceleration vector (give the x and y components). *T is the tension in the string and Fg is the gravitational force.

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The mass of the ball is 0.050kg, and at the instant shown the value of the tension is 4N. Find the net force on the ball and find the acceleration vector (give the x and y components). *T is the tension in the string and Fg is the gravitational force.

Physics

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anonymous

5 years ago

hi abby

anonymous

5 years ago

We need to better understand the orientation of the system.
Is the object being pull across the ground, it the object hanging, is the object rotation? If it is rotating, what plane is the rotation in? Horizontal or vertical?

srinidhijas method using the T-mg=ma seems to make sense, and seems like it would be correct. Eashmore I don't understand your method completely, if you could try to explain it a bit more that might help.

anonymous

5 years ago

The Force vector and the acceleration vector will have the same direction. Therefore, \[F = a_n \cdot m = \sqrt{T^2 + F_g^2}\]

anonymous

5 years ago

|dw:1332714774000:dw|

anonymous

5 years ago

T - mg = ma is not valid here.

anonymous

5 years ago

yes u did not said earlier tht it was swingin in a circle atsrule

anonymous

5 years ago

alright, sorry for not making that clear. eashmore the last equation you posted makes sense.

anonymous

5 years ago

In the x-direction, we have a tension force and only a tension force. T = 4N
In the y-direction, we have a gravitational force and only a gravitational force. Fg = 0.005 * 9.81.
The magnitude of these two forces is found from Pythagorus theorem\[F = \sqrt{~ T^2 + F_g^2} = \sqrt{~ 4^2 + (0.05 \cdot 9.81)^2}\]
To find the acceleration in each direction, realize that the tension force is caused by the centripetal force. \[T = m \cdot a_x\]and that the gravitational force causes an acceleration in the y-direction\[F_g = m \cdot a_y = m \cdot g\]
We can use these two equations to find the acceleration in the x and y-directions. Then, using the same methodology as we did to find the net force, we can find the net acceleration as\[a = \sqrt{~ a_x^2 + g^2}\]